To determine the theoretical mass of sodium sulfate produced when 355 grams of sulfuric acid reacts with an excess of sodium hydroxide, we need to follow a series of steps involving stoichiometry and conversions between mass and moles.
### Step-by-Step Solution:
1. Determine the molar mass of H₂SO₄ (Sulfuric Acid):
- The molar masses of the elements are:
- H (Hydrogen) = 1.008 g/mol
- S (Sulfur) = 32.06 g/mol
- O (Oxygen) = 16.00 g/mol
- The molar mass of H₂SO₄ is calculated as follows:
[tex]\[
\text{Molar mass of H₂SO₄} = (2 \times 1.008) + 32.06 + (4 \times 16.00) = 98.076 \text{ g/mol}
\][/tex]
2. Calculate the number of moles of H₂SO₄:
- Given mass of H₂SO₄ is 355 grams.
- Number of moles of H₂SO₄ ([tex]\(n\)[/tex]) is calculated as:
[tex]\[
n = \frac{\text{mass}}{\text{molar mass}} = \frac{355 \text{ g}}{98.076 \text{ g/mol}} = 3.619641910355235 \text{ moles}
\][/tex]
3. Determine the molar mass of Na₂SO₄ (Sodium Sulfate):
- The molar masses of the elements are:
- Na (Sodium) = 22.99 g/mol
- S (Sulfur) = 32.06 g/mol
- O (Oxygen) = 16.00 g/mol
- The molar mass of Na₂SO₄ is calculated as follows:
[tex]\[
\text{Molar mass of Na₂SO₄} = (2 \times 22.99) + 32.06 + (4 \times 16.00) = 142.04 \text{ g/mol}
\][/tex]
4. Use the stoichiometric ratio from the balanced equation to find the moles of Na₂SO₄ produced:
- The balanced equation is:
[tex]\[
2 \text{ NaOH} + \text{H₂SO₄} \rightarrow \text{Na₂SO₄} + 2 \text{H₂O}
\][/tex]
- According to the balanced equation, 1 mole of H₂SO₄ produces 1 mole of Na₂SO₄.
- Therefore, the number of moles of Na₂SO₄ produced is equal to the number of moles of H₂SO₄ reacted:
[tex]\[
3.619641910355235 \text{ moles of Na₂SO₄}
\][/tex]
5. Calculate the mass of Na₂SO₄ produced:
- Mass of Na₂SO₄ is calculated as:
[tex]\[
\text{Mass} = \text{number of moles} \times \text{molar mass} = 3.619641910355235 \text{ moles} \times 142.04 \text{ g/mol} = 514.1339369468575 \text{ g}
\][/tex]
### Final Answer:
The theoretical mass of sodium sulfate produced is approximately [tex]\(\boxed{514 \text{ grams}}\)[/tex].
